NCERT Solutions for Class 12 Maths Chapter 7 - Integrals Exercise 7.5

Exercise 7.5

1. Integrate $\dfrac{x}{{(x + 1)(x + 2)}}$

Ans: Let $\dfrac{x}{{(x + 1)(x + 2)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}}$

$ \Rightarrow x = A(x + 2) + B(x + 1)$

Equating the coefficients of ${\text{x}}$ and constant term, we obtain $A + B = 1$

$2A + B = 0$

On solving. we obtain ${\text{A}} =  - 1$ and ${\text{B}} = 2$

$\therefore \dfrac{x}{{(x + 1)(x + 2)}} = \dfrac{{ - 1}}{{(x + 1)}} + \dfrac{2}{{(x + 2)}}$

$ \Rightarrow \int {\dfrac{x}{{(x + 1)(x + 2)}}} dx = \int {\dfrac{{ - 1}}{{(x + 1)}}}  + \dfrac{2}{{(x + 2)}}dx$

$ =  - \log |x + 1| + 2\log |x + 2| + C$

$ = \log {(x + 2)^2} - \log |x + 1| + C$

$ = \log \dfrac{{{{(x + 2)}^2}}}{{(x + 1)}} + C$

Where $C$ is an arbitrary constant

2. Integrate $\dfrac{1}{{{x^2} - 9}}$

Ans: Let $\dfrac{1}{{(x + 3)(x - 3)}} = \dfrac{A}{{(x + 3)}} + \dfrac{B}{{(x - 3)}}$

$1 = A(x - 3) + B(x + 3)$

Equating the coefficients of $x$ and constant term, we obtain $A + B = 0$

$ - 3A + 3B = 1$

On solving. we obtain

${\text{A}} =  - \dfrac{1}{6}$ and ${\text{B}} = \dfrac{1}{6}$

$\dfrac{1}{{(x + 3)(x - 3)}} = \dfrac{{ - 1}}{{6(x + 3)}} + \dfrac{1}{{6(x - 3)}}$

$ \Rightarrow \int {\dfrac{1}{{\left( {{x^2} - 9} \right)}}} dx = \int {\left( {\dfrac{{ - 1}}{{6(x + 3)}} + \dfrac{1}{{6(x - 3)}}} \right)} dx$

$ =  - \dfrac{1}{6}\log |x + 3| + \dfrac{1}{6}\log |x - 3| + C$

$ = \dfrac{1}{6}\log \dfrac{{|(x - 3)|}}{{|(x + 3)|}} + C$

Where $C$ is an arbitrary constant

3. Integrate $\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}}$

Ans: Let $\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}$

$3x - 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B + C = 0$

$ - 5A - 4B - 3C = 3$

$6A + 3B + 2C =  - 1$

Solving these equations, we obtain ${\text{A}} = 1,\;{\text{B}} =  - 5$, and $C = 4$

$\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{1}{{(x - 1)}} - \dfrac{5}{{(x - 2)}} + \dfrac{4}{{(x - 3)}}$

$ \Rightarrow \int {\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}}} dx = \int {\left\{ {\dfrac{1}{{(x - 1)}} - \dfrac{5}{{(x - 2)}} + \dfrac{4}{{(x - 3)}}} \right\}} dx$

$ = \log |x - 1| - 5\log |x - 2| + 4\log |x - 3| + C$

Where $C$ is an arbitrary constant.

4. Integrate $\dfrac{x}{{(x - 1)(x - 2)(x - 3)}}$

Ans: $\dfrac{x}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}$

$x = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B + C = 0$

$45 - 3C = 1$

$6A + 4B + 2C = 0$

Solving these equations, we obtain $A = \dfrac{1}{2},B = 2$ and $C = \dfrac{3}{2}$

$\therefore \dfrac{x}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{1}{{2(x - 1)}} - \dfrac{2}{{(x - 2)}} + \dfrac{3}{{2(x - 3)}}$

$ \Rightarrow \int {\dfrac{x}{{(x - 1)(x - 2)(x - 3)}}} dx = \int {\left\{ {\dfrac{1}{{2(x - 1)}} - \dfrac{2}{{(x - 2)}} + \dfrac{3}{{2(x - 3)}}} \right\}} dx$

$ = \dfrac{1}{2}\log |x - 1| - 2\log |x - 2| + \dfrac{3}{2}\log |x - 3| + C$

Where $C$ is an arbitrary constant.

5. Integrate $\dfrac{{2x}}{{{x^2} + 3x + 2}}$

Ans: Let $\dfrac{{2x}}{{{x^2} + 3x + 2}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}}$

$2x = A(x + 2) + B(x + 1)$

$ \ldots (1)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B = 2$

$2A + B = 0$

Solving these equations, we obtain $A =  - 2$ and ${\mathbf{B}} = 4$

$\therefore \dfrac{{2x}}{{(x + 1)(x + 2)}} = \dfrac{{ - 2}}{{(x + 1)}} + \dfrac{4}{{(x + 2)}}$

$ \Rightarrow \int {\dfrac{{2x}}{{(x + 1)(x + 2)}}} dx = \int {\left\{ {\dfrac{4}{{(x + 2)}} - \dfrac{2}{{(x + 1)}}} \right\}} dx$

$ = 4\log |x + 2| - 2\log |x + 1| + C$

Where $C$ is an arbitrary constant.

6. Integrate $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}}$

Ans: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left( {1 - {x^2}} \right)$ by $x(1 - 2x)$, we obtain $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}} = \dfrac{1}{2} + \dfrac{1}{2}\left( {\dfrac{{2 - x}}{{x(1 - 2x)}}} \right)$

Let $\dfrac{{2 - x}}{{x(1 - 2x)}} = \dfrac{A}{x} + \dfrac{B}{{(1 - 2x)}}$

$ \Rightarrow (2 - x) = A(1 - 2x) + Bx$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $ - 2A + B =  - 1$

And $A = 2$ Solving these equations, we obtain $A = 2$ and $B = 3$ $\therefore \dfrac{{2 - x}}{{x(1 - 2x)}} = \dfrac{2}{x} + \dfrac{3}{{1 - 2x}}$

Substituting in equation (1), we obtain $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}} = \dfrac{1}{2} + \dfrac{1}{2}\left\{ {\dfrac{2}{x} + \dfrac{3}{{(1 - 2x)}}} \right\}$

$\int {\dfrac{{1 - {x^2}}}{{x(1 - 2x)}}} dx = \int {\left\{ {\dfrac{1}{2} + \dfrac{1}{2}\left( {\dfrac{2}{x} + \dfrac{3}{{(1 - 2x)}}} \right)} \right\}} dx$

$ = \dfrac{x}{2} + \log |x| + \dfrac{3}{{2( - 2)}}\log |1 - 2x| + C$

$ = \dfrac{x}{2} + \log |x| - \dfrac{3}{4}\log |1 - 2x| + c$

Where $C$ is an arbitrary constant.

7. Integrate $\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}$

Ans: Let $\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}} = \dfrac{{Ax + B}}{{\left( {{x^2} + 1} \right)}} + \dfrac{C}{{(x - 1)}}$

$x = (Ax + B)(x - 1) + C\left( {{x^2} + 1} \right)$

$x = A{x^2} - Ax + Bx - B + C{x^2} + C$

Equating the coefficients of ${x^2},x$, and constant term, we obtain

A $ - A + B = 1$

$ - B + C = 0$

On solving these equations, we obtain ${\text{A}} =  - \dfrac{1}{2},\;{\text{B}} = \dfrac{1}{2}$, and ${\text{C}} = \dfrac{1}{2}$

From equation (1), vre obtain $\therefore \dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}} = \dfrac{{\left( { - \dfrac{1}{2}x + \dfrac{1}{2}} \right)}}{{{x^2} + 1}} + \dfrac{{\dfrac{1}{2}}}{{(x - 1)}}$

$ \Rightarrow \int {\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}}  =  - \dfrac{1}{2}\int {\dfrac{x}{{{x^2} + 1}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{{x^2} + 1}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{x - 1}}} dx$

$ =  - \dfrac{1}{4}\int {\dfrac{{2x}}{{{x^2} + 1}}} dx + \dfrac{1}{2}{\tan ^{ - 1}}x + \dfrac{1}{2}\log |x - 1| + C$

Consider $\int {\dfrac{{2x}}{{{x^2} + 1}}} dx$, let $\left( {{x^2} + 1} \right) = t \Rightarrow 2xdx = dt$

$ \Rightarrow \int {\dfrac{{2x}}{{{x^2} + 1}}} dx - \int {\dfrac{{dt}}{t}}  - \log |t| - \log \left| {{x^2} + 1} \right|$

$\therefore \int {\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}}  =  - \dfrac{1}{4}\log \left| {{x^2} + 1} \right| + \dfrac{1}{2}{\tan ^{ - 1}}x + \dfrac{1}{2}\log |x - 1| + C$

$ = \dfrac{1}{2}\log |{\text{x}} - 1| - \dfrac{1}{4}\log \left| {{{\text{x}}^2} + 1} \right| + \dfrac{1}{2}{\tan ^{ - 1}}{\text{x}} + {\text{C}}$

Where $C$ is an arbitrary constant.

8. Integrate $\dfrac{x}{{{{(x - 1)}^2}(x + 2)}}$

Ans: Let $\dfrac{x}{{{{(x - 1)}^2}(x + 2)}} - \dfrac{A}{{(x - 1)}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 2)}}$

$x = A(x - 1)(x + 2) + B(x + 2) + C{(x - 1)^2}$

Equating the coefficients of ${{\text{x}}^2},{\text{x}}$ and constant term, we obtain ${\text{A}} + {\text{C}} = 0$

$A + B - 2C = 1$

$ - 2\;{\text{A}} + 2\;{\text{B}} + {\text{C}} = 0$

On solving. we obtain $A = \dfrac{2}{9}$ and $C = \dfrac{{ - 2}}{9}$

${\text{B}} = \dfrac{1}{3}$

$\therefore \dfrac{x}{{{{(x - 1)}^2}(x + 2)}} = \dfrac{2}{{9(x - 1)}} + \dfrac{1}{{3{{(x - 1)}^2}}} - \dfrac{2}{{9(x + 2)}}$

$ \Rightarrow \int {\dfrac{x}{{{{(x - 1)}^2}(x + 2)}}} dx - \dfrac{2}{9}\int {\dfrac{1}{{(x - 1)}}} dx + \dfrac{1}{3}\int {\dfrac{1}{{{{(x - 1)}^2}}}} dx - \dfrac{2}{9}\int {\dfrac{1}{{(x + 2)}}} dx$

$ = \dfrac{2}{9}\log |x - 1| + \dfrac{1}{3}\left( {\dfrac{{ - 1}}{{x - 1}}} \right) - \dfrac{2}{9}\log |x + 2| + C$

$ = \dfrac{2}{9}\log \left| {\dfrac{{x - 1}}{{x + 2}}} \right| - \dfrac{1}{{3(x - 1)}} + C$

Where $C$ is an arbitrary constant.

9. Integrate $\dfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}}$

Ans: $\dfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}} = \dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}}$

Let $\dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 1)}}$

$3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C{(x - 1)^2}$

$3x + 5 = A\left( {{x^2} - 1} \right) + B(x + 1) + C\left( {{x^2} + 1 - 2x} \right)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + C = 0$

$B - 2C - 3$

$ - A + B + C = 5$

On solving. we obtain $B = 4$

${\text{A}} =  - \dfrac{1}{2}$ and ${\text{C}} = \dfrac{1}{2}$

$\therefore \dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}} = \dfrac{{ - 1}}{{2(x - 1)}} + \dfrac{4}{{{{(x - 1)}^2}}} + \dfrac{1}{{2(x + 1)}}$

$ \Rightarrow \int {\dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}}} dx =  - \dfrac{1}{2}\int {\dfrac{1}{{x - 1}}} dx + 4\int {\dfrac{1}{{{{(x - 1)}^2}}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{(x + 1)}}} dx$

$ =  - \dfrac{1}{2}\log |x - 1| + 4\left( {\dfrac{{ - 1}}{{x - 1}}} \right) + \dfrac{1}{2}\log |x + 1| + C$

$ = \dfrac{1}{2}\log \left| {\dfrac{{x + 1}}{{x - 1}}} \right| - \dfrac{4}{{(x - 1)}} + C$

Where $C$ is an arbitrary constant.

10. Integrate $\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}}$

Ans: $\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}} = \dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}}$

Let $\dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}} + \dfrac{C}{{(2x + 3)}}$

$ \Rightarrow (2x - 3) = A(x - 1)(2x + 3) + B(x + 1)(2x + 3) + C(x + 1)(x - 1)$

$ \Rightarrow (2x - 3) = A\left( {2{x^2} + x - 3} \right) + B\left( {2{x^2} + 5x + 3} \right) + C\left( {{x^2} - 1} \right)$

$ \Rightarrow (2x - 3) = (2A + 2B + C){x^2} + (A + 5B)x + ( - 3A + 3B - C)$

Equating the coefficients of ${x^2},x$ and constant, we obtain $2A + 2B + C = 0$

$A + 5B = 2$

$ - 3A + 3B - C =  - 3$

On solving, we obtain ${\text{B}} =  - \dfrac{1}{{10}},\;{\text{A}} = \dfrac{5}{2}$, and ${\text{C}} =  - \dfrac{{24}}{5}$

$\therefore \dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}} = \dfrac{5}{{2(x + 1)}} - \dfrac{1}{{10(x - 1)}} - \dfrac{{24}}{{5(2x + 3)}}$

$ \Rightarrow \int {\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}}} dx = \dfrac{5}{2}\int {\dfrac{1}{{(x + 1)}}} dx - \dfrac{1}{{10}}\int {\dfrac{1}{{x - 1}}} dx - \dfrac{{24}}{5}\int {\dfrac{1}{{(2x + 3)}}} dx$

$ = \dfrac{5}{2}\log |x + 1| - \dfrac{1}{{10}}\log |x - 1| - \dfrac{{24}}{{5 \times 2}}\log |2x + 3|$

$ = \dfrac{5}{2}\log |x + 1| - \dfrac{1}{{10}}\log |x - 1| - \dfrac{{12}}{5}\log |2x + 3| + C$

Where $C$ is an arbitrary constant.

11. Integrate $\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}}$

Ans : $\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}} = \dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}}$

Let $\dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}} + \dfrac{C}{{(x - 2)}}$

$5x = A(x + 2)(x - 2) + B(x + 1)(x - 2) + C(x + 1)(x + 2)$

Equating the coefficients of ${x^2},x$ and constant, we obtain $A + B + C = 0$

$B + 3C = 5$ and $4A - 2B + 2C = 0$

On solving. we obtain $A - \dfrac{5}{3},B -  - \dfrac{5}{2}$, and $C - \dfrac{5}{6}$

$\therefore \dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}} = \dfrac{5}{{3(x + 1)}} +  - \dfrac{5}{{2(x + 2)}} + \dfrac{5}{{6(x - 2)}}$

$ \Rightarrow \int {\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}}} dx = \dfrac{5}{3}\int {\dfrac{1}{{(x + 1)}}} dx - \dfrac{5}{2}\int {\dfrac{1}{{(x + 2)}}} dx + \dfrac{5}{6}\int {\dfrac{1}{{(x - 2)}}} dx$

$ = \dfrac{5}{3}\log |x + 1| - \dfrac{5}{2}\log |x + 2| + \dfrac{5}{6}\log |x - 2| + C$

Where $C$ is an arbitrary constant.

12. Integrate $\dfrac{{{x^3} + x + 1}}{{{x^2} - 1}}$

Ans: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left( {{x^3} + x + 1} \right)$ by ${x^2} - 1$, we obtain $\dfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \dfrac{{2x + 1}}{{{x^2} - 1}}$

Let $\dfrac{{2x + 1}}{{{x^2} - 1}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}}$

$2x + 1 = A(x - 1) + B(x + 1)$

Equating the coefficients of $x$ and constant, we obtain $A + B = 2$

$ - A + B = 1$

On solving. we obtain $A - \dfrac{1}{2}$ and $B - \dfrac{3}{2}$

$\therefore \dfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \dfrac{1}{{2(x + 1)}} + \dfrac{3}{{2(x - 1)}}$

$ \Rightarrow \int {\dfrac{{{x^3} + x + 1}}{{{x^2} + 1}}} dx = \int x dx + \dfrac{1}{2}\int {\dfrac{1}{{(x + 1)}}} dx + \dfrac{3}{2}\int {\dfrac{1}{{(x - 1)}}} dx$

$ = \dfrac{{{x^2}}}{2} + \log |x + 1| + \dfrac{3}{2}\log |x - 1| + C$

Where $C$ is an arbitrary constant.

13. Integrate $\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}}$

Ans:

Let $\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}} = \dfrac{A}{{(1 - x)}} + \dfrac{{Bx + C}}{{\left( {1 + {x^2}} \right)}}$

$2 = A\left( {1 + {x^2}} \right) + (Bx + C)(1 - x)$

$2 = A + A{x^2} + Bx - B{x^2} + C - Cx$

Equating the coefficient of ${x^2},x$, and constant term, we obtain ${\text{A}} - {\text{B}} = 0$

${\mathbf{B}} - {\mathbf{C}} = {\mathbf{0}}$

$A + C = 2$

On solving these equations, we obtain $A = 1,B = 1$, and $C = 1$

$\therefore \dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}} = \dfrac{1}{{1 - x}} + \dfrac{{x + 1}}{{1 + {x^2}}}$

$ \Rightarrow \int {\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}}} dx = \int {\dfrac{1}{{1 - x}}} dx + \int {\dfrac{x}{{1 + {x^2}}}} dx + \int {\dfrac{1}{{1 + {x^2}}}} dx$

$ =  - \int {\dfrac{1}{{1 - x}}} dx + \dfrac{1}{2}\int {\dfrac{{2x}}{{1 + {x^2}}}} dx + \int {\dfrac{1}{{1 + {x^2}}}} dx$

$ =  - \log |x - 1| + \dfrac{1}{2}\log \left| {1 + {x^2}} \right| + {\tan ^{ - 1}}x + C$

Where $C$ is an arbitrary constant.

14. Integrate $\dfrac{{3x - 1}}{{{{(x + 2)}^2}}}$

Ans:

Let $\dfrac{{3x - 1}}{{{{(x + 2)}^2}}} = \dfrac{A}{{(x + 2)}} + \dfrac{B}{{{{(x + 2)}^2}}}$

$ \Rightarrow 3x - 1 = A(x + 2) + B$

Equating the coefficient of $x$ and constant term, we obtain $A = 3$

$2A + B =  - 1 \Rightarrow \mathbb{B} -  - 7$

$\therefore \dfrac{{3x - 1}}{{{{(x + 2)}^2}}} = \dfrac{3}{{(x + 2)}} - \dfrac{7}{{{{(x + 2)}^2}}}$

$ \Rightarrow \int {\dfrac{{3x - 1}}{{{{(x + 2)}^2}}}} dx = 3\int {\dfrac{1}{{(x + 2)}}} dx - 7\int {\dfrac{1}{{{{(x + 2)}^2}}}} dx$

$ = 3\log |x + 2| - 7\left( {\dfrac{{ - 1}}{{(x + 2)}}} \right) + C$

$ = 3\log |x + 2| + \dfrac{7}{{(x + 2)}} + C$

Where $C$ is an arbitrary constant.

15. Integrate $\dfrac{1}{{{x^4} - 1}}$

Ans :

$\dfrac{1}{{\left( {{x^4} - 1} \right)}} - \dfrac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}} - \dfrac{1}{{(x + 1)(x - 1)\left( {1 + {x^2}} \right)}}$

Let $\dfrac{1}{{(x + 1)(x - 1)\left( {1 + {x^2}} \right)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + 1} \right)}}$

$1 = A(x - 1)\left( {1 + {x^2}} \right) + B(x + 1)\left( {1 + {x^2}} \right) + (Cx + D)\left( {{x^2} - 1} \right)$

$1 = A\left( {{x^3} + x - {x^2} - 1} \right) + B\left( {{x^3} + x + {x^2} + 1} \right) + C{x^3} + D{x^2} - Cx - D$

$1 = (A + B + C){x^3} + ( - A + B + D){x^2} + (A + B - C)x + ( - A + B - D)$

Equating the coefficient of ${x^3},{x^2},x$, and constant term, we obtain $A + B + C = 0$

$ - A + B + D = 0$

$A + B - C = 0$

$ - A + B - D = 1$

${\text{A}} =  - \dfrac{1}{4},\;{\text{B}} = \dfrac{1}{4},{\text{C}} = {\text{O}}$, and ${\text{D}} =  - \dfrac{1}{2}$

$\therefore \dfrac{1}{{\left( {{x^n} - 1} \right)}} = \dfrac{{ - 1}}{{4(x + 1)}} + \dfrac{1}{{4(x - 1)}} + \dfrac{1}{{2\left( {{x^2} + 1} \right)}}$

$ \Rightarrow \int {\dfrac{1}{{{x^4} - 1}}} dx -  - \dfrac{1}{4}\log |x - 1| + \dfrac{1}{4}\log |x - 1| - \dfrac{1}{2}{\tan ^1}x + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| - \dfrac{1}{2}{\tan ^1}x + C$

Where $C$ is an arbitrary constant.

16. Integrate $\dfrac{1}{{x\left( {{x^n} + 1} \right)}}$

Hint: multiply numerator and denominator by ${x^{n - 1}}$ and put $x^n = t$

Ans:

 $\dfrac{1}{{x\left( {{x^n} + 1} \right)}}$

Multiplying numerator and denominator by ${x^{n - 1}}$, we obtain $\dfrac{1}{{x\left( {{x^n} + 1} \right)}} = \dfrac{{{x^{n - 1}}}}{{{x^{n - 1}}x\left( {{x^n} + 1} \right)}} = \dfrac{{{x^{n - 1}}}}{{{x^n}\left( {{x^n} + 1} \right)}}$

Let ${x^n} = t \Rightarrow n{x^{ - 1}}dx = dt$

$\therefore \int {\dfrac{1}{{x\left( {{x^n} + 1} \right)}}} dx = \int {\dfrac{{{x^{n - 1}}}}{{{x^7}\left( {{x^n} + 1} \right)}}} dx = \dfrac{1}{n}\int {\dfrac{1}{{t(t + 1)}}} dt$

Let $\dfrac{1}{{t(t + 1)}} = \dfrac{A}{t} + \dfrac{B}{{(t + 1)}}$

$1 = A(1 + t) + Bt$

Equating the coefficients of $t$ and constant, ve obtain $A = 1$ and $B =  - 1$

$\dfrac{1}{{t(t + 1)}} = \dfrac{1}{t} - \dfrac{1}{{(1 + t)}}$

$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^4} + 1} \right)}}} dx = \dfrac{1}{n}\int {\left\{ {\dfrac{1}{t} - \dfrac{1}{{(1 + t)}}} \right\}} dx$

$\dfrac{1}{n}[\log |t| - \log |t + 1|] + C$

$ = \dfrac{1}{n}\left[ {\log \left| {{x^n}} \right| - \log \left| {{x^n} + 1} \right|} \right] + C$

$ = \dfrac{1}{n}\log \left| {\dfrac{{{x^n}}}{{{x^2} + 1}}} \right| + C$

Where $C$ is an arbitrary constant.

17. Integrate $\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$

(Hint: Put $\sin x = t]$

Ans:

$\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$

let $\sin x = t \Rightarrow \cos xdx = dt$

$\therefore \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} dx = \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}} $

Let $\dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{A}{{(1 - t)}} + \dfrac{B}{{(2 - t)}}$

$1 = A(2 - t) + B(1 - t)$

Equating the coefficients of $t$ and constant, wre obtain $ - A - B = 0$ and $2A + B = 1$

On solving. we obtain $A = 1$ and $B =  - 1$

$\therefore \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{1}{{(1 - t)}} - \dfrac{1}{{(2 - t)}}$

$ \Rightarrow \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} dx = \int {\left\{ {\dfrac{1}{{1 - t}} - \dfrac{1}{{(2 - t)}}} \right\}} dt$

$ = \log \left| {\dfrac{{2 - t}}{{1 - t}}} \right| + C$

$ = \log \left| {\dfrac{{2 - \sin x}}{{1 - \sin x}}} \right| + C$

Where $C$ is an arbitrary constant.

18. Integrate $\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}$

Ans:

$\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \dfrac{{4{x^2} + 10}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}$

Let $\dfrac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{Ax + B}}{{\left( {{x^2} + 3} \right)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + 4} \right)}}$

$4{x^2} + 10 = (Ax + B)\left( {{x^2} + 4} \right) + (Cx + D)\left( {{x^2} + 3} \right)$

$4{x^2} + 10 = A{x^3} + 4Ax + B{x^2} + 4B + C{x^3} + 3Cx + D{x^2} + 3D$

$4{x^2} + 10 = (A + C){x^2} + (B + D){x^2} + (4A + 3C)x + (4B + 3D)$

Equating the coefficients of ${x^3},{x^2},x$ and constant term, we obtain $A + C = 0$

$B + D = 4$

$4A + 3C = 0$

$4B + 3D = 10$

On solving these equations, we obtain ${\text{A}} = 0,\;{\text{B}} =  - 2,{\text{C}} = 0$, and ${\text{D}} = 6$

$\therefore \dfrac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \dfrac{6}{{\left( {{x^2} + 4} \right)}}$

$\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \left( {\dfrac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \dfrac{6}{{\left( {{x^2} + 4} \right)}}} \right)$

$ \Rightarrow \int {\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}} dx = \int {\left\{ {1 + \dfrac{2}{{\left( {{x^2} + 3} \right)}} - \dfrac{6}{{\left( {{x^2} + 4} \right)}}} \right\}} dx$

$ = \int {\left\{ {1 + \dfrac{2}{{{x^2} + {{(\sqrt 3 )}^2}}} - \dfrac{6}{{{x^2} + {2^2}}}} \right\}} $

$ = x + 2\left( {\dfrac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\dfrac{x}{{\sqrt 3 }}} \right) - 6\left( {\dfrac{1}{2}{{\tan }^{ - 1}}\dfrac{x}{2}} \right) + C$

$ = x + \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{x}{{\sqrt 3 }} - 3{\tan ^{ - 1}}\dfrac{x}{2} + C$

Where $C$ is an arbitrary constant.

19. Integrate $\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Ans:

$\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Let ${x^2} = t \Rightarrow 2xdx = dt$

$\therefore \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\dfrac{{dt}}{{(t + 1)(t + 3)}}} $

Let $\dfrac{1}{{(t + 1)(t + 3)}} = \dfrac{A}{{(t + 1)}} + \dfrac{B}{{(t + 3)}}$

$1 = A(t + 3) + B(t + 1)$

Equating the coefficients of ${\text{t}}$ and constant, we obtain $A + B = 0$ and $3A + B = 1$

On solving. we obtain ${\text{A}} = \dfrac{1}{2}$ and ${\text{B}} =  - \dfrac{1}{2}$

$\therefore \dfrac{1}{{(t + 1)(t + 3)}} = \dfrac{1}{{2(t + 1)}} - \dfrac{1}{{2(t + 3)}}$

$ \Rightarrow \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\left\{ {\dfrac{1}{{2(t + 1)}} - \dfrac{1}{{2(t + 3)}}} \right\}} dt$

$ = \dfrac{1}{2}\log |(t + 1)| - \dfrac{1}{2}\log |t + 3| + C$

$ = \dfrac{1}{2}\log \left| {\dfrac{{t + 1}}{{t + 3}}} \right| + C$

$ = \dfrac{1}{2}\log \left| {\dfrac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + C$

Where $C$ is an arbitrary constant.

20. Integrate $\dfrac{1}{{x\left( {{x^4} - 1} \right)}}$

Ans:

$\dfrac{1}{{x\left( {{x^4} - 1} \right)}}$

Multiplying numerator and denominator by ${x^3}$, we obtain

$\dfrac{1}{{x\left( {{x^4} - 1} \right)}} = \dfrac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}$

$\therefore \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \int {\dfrac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}} dx$

Let ${x^4} = t \Rightarrow 4{x^3}dx = dt$

$\therefore \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \dfrac{1}{4}\int {\dfrac{{dt}}{{t(t - 1)}}} $

Let $\dfrac{1}{{t(t - 1)}} = \dfrac{A}{t} + \dfrac{B}{{(t - 1)}}$

$1 = {\text{A}}({\text{t}} - 1) + {\text{Bt}}$

Equating the coefficients of $t$ and constant, we obtain $A + B = 0$ and $ - A = 1$

${\text{A}} =  - 1$ and ${\mathbf{B}} = 1$

$ \Rightarrow \dfrac{1}{{t(t - 1)}} = \dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}$

$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \dfrac{1}{4}\int {\left\{ {\dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}} \right\}} dt$

$ - \dfrac{1}{4}[ - \log |t| + \log |t - 1|] + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{t - 1}}{t}} \right| + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{{x^4} - 1\mid }}{{{x^4}}}} \right| + C$

Where $C$ is an arbitrary constant.

21. Integrate $\dfrac{1}{{\left( {{{\text{e}}^ x } - 1} \right)}}$

Hint: Put $\dfrac{1}{e^x - 1}$

Ans :

Let ${e^x} = t \Rightarrow {e^x}dx = dt$

$ \Rightarrow \int {\dfrac{1}{{\left( {{{\text{e}}^x} - 1} \right)}}} {\text{dx}} = \int {\dfrac{1}{{{\text{t}} - 1}}}  \times \dfrac{{{\text{dt}}}}{{\text{t}}} - \int {\dfrac{1}{{{\text{t}}({\text{t}} - 1)}}} {\text{dt}}$

Let $\dfrac{1}{{t(t - 1)}} = \dfrac{A}{t} + \dfrac{B}{{t - 1}}$

$1 = {\text{A}}({\text{t}} - 1) + {\text{Bt}}$

Equating the coefficients of ${\text{t}}$ and constant, we obtain

$A + B = 0$ and $ - A = 1$

$A =  - 1$ and ${\mathbf{B}} = {\mathbf{1}}$

$\therefore \dfrac{1}{{t(t - 1)}} = \dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}$

$ \Rightarrow \int {\dfrac{1}{{t(t - 1)}}} dt = \log \left| {\dfrac{{t - 1}}{t}} \right| + C$

$ = \log \left| {\dfrac{{{{\text{e}}^x} - 1}}{{{{\text{e}}^x}}}} \right| + {\text{C}}$

Where $C$ is an arbitrary constant.

22. $\int {\dfrac{{xdx}}{{(x - 1)(x - 2)}}} $ equals

a. $A\log \left| {\dfrac{{{{(x - 1)}^2}}}{{x - 2}}} \right| + C$

b. $\log \left| {\dfrac{{{{(x - 2)}^2}}}{{x - 1}}} \right| + C$

c. $\log \left| {{{\left( {\dfrac{{x - 1}}{{x - 2}}} \right)}^2}} \right| + C$

d. $\log |(x - 1)(x - 2)| + C$

Ans:

Let $\dfrac{x}{{(x - 1)(x - 2)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}}$

$x = A(x - 2) + B(x - 1)$

Equating the coefficients of $x$ and constant, we obtain $A + B = 1$ and $ - 2A - B = 0$

$A--1$ and $B = 2$

$\dfrac{x}{{(x - 1)(x - 2)}} =  - \dfrac{1}{{(x - 1)}} + \dfrac{2}{{(x - 2)}}$

$ \Rightarrow \int {\dfrac{x}{{(x - 1)(x - 2)}}} dx = \int {\left\{ {\dfrac{{ - 1}}{{(x - 1)}} + \dfrac{2}{{(x - 2)}}} \right\}} dx$

$ =  - \log |x - 1| + 2\log |x - 2| + C$

$ = \log \left| {\dfrac{{{{(x - 2)}^2}}}{{x - 1}}} \right| + C$

Hence, the correct Answer is $B$.

23. $\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}} $ equals

a. $\log |x| - \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

b. $\log |x| + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

c. $ - \log |x| + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

d. $\dfrac{1}{2}\log |x| + \log \left( {{x^2} + 1} \right) + C$

Ans :

Let $\dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{A}{x} + \dfrac{{Bx + C}}{{{x^2} + 1}}$

$1 = A\left( {{x^2} + 1} \right) + (Bx + C)x$

Equating the coefficients of ${x^2},x$, and constant term, we obtain $A + B = 0$

$c = 0$

$A = 1$

On solving these equations, we obtain $A = 1,B =  - 1$, and $C = 0$

$\therefore \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{1}{x} + \dfrac{{ - x}}{{{x^2} + 1}}$

$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^2} + 1} \right)}}} dx = \int {\left\{ {\dfrac{1}{x} - \dfrac{x}{{{x^2} + 1}}} \right\}} dx$

$ - \log |x| - \dfrac{1}{2}\log \left| {{x^2} + 1} \right| + C$

Hence, the correct Answer is ${\text{A}}$. 

Where $C$ is an arbitrary constant.

Alter:

$ \Rightarrow \int {\dfrac{1}{{{\text{x}}\left( {{{\text{x}}^2} + 1} \right)}}} dx = \int {\left\{ {\dfrac{{\text{x}}}{{{{\text{x}}^2}\left( {{{\text{x}}^2} + 1} \right)}}} \right\}} {\text{dx}}$

Let ${x^2} = t$, therefore, $2xdx = dt$ $\therefore \int {\dfrac{x}{{{x^2}\left( {{x^2} + 1} \right)}}} dx = \dfrac{1}{2}\int {\dfrac{{dt}}{{t(t + 1)}}}  = \dfrac{1}{2}\int {\dfrac{{(t + 1) - t}}{{t(t + 1)}}} dt = \dfrac{1}{2}\int \frac{1}{t} -  \dfrac{1}{{t + 1}}dt$

$ - \dfrac{1}{2}[\log t - \log (t + 1)] + C$

$ = \log |x| - \dfrac{1}{2}\log \left| {{x^2} + 1} \right| + C$

Where $C$ is an arbitrary constant.

Conclusion

The NCERT Solutions for Class 12 Chapter 7 Integrals Exercise 7.5 focuses on the application of integration techniques to solve various types of integral problems, particularly focusing on integration by parts and the integration of rational functions. It's essential to understand the fundamental integration formulas and the steps for applying integration by parts, as these are commonly used methods in this exercise. Emphasizing practice on these techniques will help in solving the problems efficiently.

Class 12 Maths Chapter 7: Exercises Breakdown

CBSE Class 12 Maths Chapter 7 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

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